Val kotlin_toLong10Radix = "10". Val kotlin_toLong010Radix = "010".toLong(8) // 8 as "octal" parsing is forced Val java_parseLong10 = ("10") // 10 as parsed as decimal When converting a higher datatype to lower datatype, it is. Val java_parseLong010 = ("010") // 10 as parsed as decimal Long is a larger data type than int, we need to explicitly perform typecasting for the conversion. Val kotlin_toLong10 = "10".toLong() // 10 as parsed as decimal val kotlin_toLong010 = "010".toLong() // 10 as parsed as decimal That means that decode can parse Strings like "0x412", where other methods will result in a NumberFormatException. (Sign) DecimalNumeral | (Sign) 0x HexDigits | (Sign) 0X HexDigits | (Sign) # HexDigits | (Sign) 0 OctalDigits Numbers given by the following grammar: DecodableString: Īnd here comes (String) into the picture:ĭecodes a String into a Long. The characters in the string must all be digits of the specified radix. Parses the string argument as a signed long in the radix specified by Kotlin's String.toLong(radix: Int) is equivalent to Java's e Long.parseLong(String, int): Radix 10 were given as arguments to the parseLong(, int) method. Resulting long value is returned, exactly as if the argument and the Parses the string argument as a signed decimal long. Kotlin's String.toLong() is equivalent to Java's Long.parseLong(String): One good old Java possibility what's not mentioned in the answers is (String). (string) public static Long valueOf(String s) throws NumberFormatException public fun String.toLongOrNull(radix: Int): Long? ĥ. If the string is not a valid representation of a IllegalArgumentException when is not a valid radix for string public inline fun String.toLong(radix: Int): Long = (this, checkRadix(radix)) is not a valid radix for string to number conversion. Representation of a IllegalArgumentException when If the string is not a valid representation of a number. Parses the string as a number and returns the result or null If you wanted to test that you can parse a string, you could implement a tryParse.Parses the string as a number and returns the NumberFormatException if the string is not a valid The Exceptions topic discusses how to deal with such exceptions. The numeric conversions above will all throw an (unchecked) NumberFormatException if you attempt to parse a string that is not a suitably formatted number, or is out of range for the target type. String to BigDecimal: String bigFraction = "17.21455" īigDecimal reallyBig = new BigDecimal(bigFraction) String to BigInteger: String bigNumber = "21" īigInteger reallyBig = new BigInteger(bigNumber) String to boolean: String falseString = "False" īoolean falseBool = Boolean.parseBoolean(falseString) // falseBool = falseīoolean trueBool = Boolean.parseBoolean(trueString) // trueBool = true String to double: String double = "1.47" You can convert a numeric string to various Java numeric types as follows:
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